I know. That makes the hunger for more results stronger... It is like you want your favorite candy, it is so within reach... then you can finally grab it.
Then it tastes much better right?
And if you have a hangover it is actually the best ever tasted.
The hangover is the analogy for the failure.
Holy cow, I am producing so much nonsense this evening! Even more than last week.
@JonasTeuwen yes i remember research. once i was given a project by my undergrad research advisor. i spent months showing a result only to find i had been scooped.
THEOREM 1.21 Let $H$ be a lineal homogeneous sistem of $n$ equations and $m$ unknowns, then if $n<m$, there exists $X \in K^m, X\neq 0$, which is a solution of the system.
Shouldn't it be $n>m$?
If there are more unknowns than equations shouldn't that be a problem?
seriously, the lattice of subgroups should tell us that bigger subgroups are "higher" than smaller ones. yet the upper cs starts at the bottom and the lower cs starts at the top.
Due to my limited math knowledge, I am rather used to not knowing most of the terminology here, and so when i saw that i thought its one of them, but after a moment when i figured out, Oh! man even i understand this.
@anon I just used it, but I don't see him use iether, since little importance is given to that group for he moves on to Characters right away, it seems.
What is a bounty?
What is the "Featured" tab on the homepage?
How can I search for questions that have a bounty attached?
How do I start a bounty? When can I start a bounty?
How long is the bounty period?
How do I award a bounty?
Can I award a bounty to my own answer?
What happens if there's no ...
The $o(f(n,k))$ means a function $g(n,k)$ such that for all $\epsilon>0$, we have $|g(n,k)| \le \epsilon|f(n,k)|$ holds for all $1\le k\le n$ and $n\ge N_\epsilon$.
The main difference between $O$ and $o$ is whether the constant in $O$ or $o$ notation is related to the variables. For example, $1/(n-1)=O(1/n)$ as $n\to\infty$ means that there exists $M$ and $n_0$ such that $|1/(n-1)|\le M|1/n|$ holds for all $n>n_0$, where $n_0$ is somewhat unrelated to $M$, but $1/(n-1)=o(1)$ as $n\to\infty$ means that for all $\epsilon>0$, there exists $N_\epsilon$ such that $|1/(n-1)|\le \epsilon$ holds for all $n>N_\epsilon$.
I wonder a systematical way to deal with limits just like algebra but not calculus.
Because I find that L'Hospital's rule for $\infty/\infty$ at $x\to a$ is not easy to prove.
Just like proving SC.
And I find that most calculus books more often uses $o$-notation than $O$-notation, but I find out that $o$-notation is more hard to manipulate, or I have not got the clear idea of $o$-notation?
Notice that $\sum_{k=1}^nO(f(n,k))=O(\sum_{k=1}^n|f(n,k)|)$ is much easier than the one for $o$.
FINAL EDIT : Prove that if $p^z|n^2-1$
$$p^{x-z}(p^{z}-1)=\dfrac{ n^2-1}{p^z}-3$$ doesn't hold for any chosen values of $p,x,n$ and $z$.
Here $p>3$ is an odd prime , $x=2y+z, \ \{\{x,y,z\}>0\} \in \mathbb{Z}$ . There $n$ is an even number.
If the above statement is prove it will lea...
@lyengar Please avoid using sir online. I've asked a person and he said that it's seldom used online, especially when you don't know the gender of whom you're talking to.
@FrankScience : But sir, I am just saying that out of respect.. I know the people whom I am talking with are men..., I have high gratitude towards everyone so I used that sir..
$A\iff B\iff C$ is ambiguous, doesn't mean there isn't a conventional interpretation of it as $A\iff B$ and $B\iff C$ (in contexts that aren't pure logic anyway)
@leo The rigorous interpretation is the set $$\left\{\;\sum_{k=1}^ng(n,k)\;\bigg\vert\;|g(n,k)|\le M|f(n,k)|\hbox{ holds for some $M$ and over all $1\le k\le n$}\;\right\}$$
the number of terms is actually a function of $n$; such is not the case in the latter and so induction will not work (induction will only reach fixed numbers of summands)
Yes, I think so too. Or at least some basic $o$-machinery developed first with $\epsilon$-$N$s.
@JonasTeuwen I never claimed that. But I guess that those MO threads I found will be much more useful than anything I would be able to tell you about that problem.
@MartinSleziak Oh, I don't want to philosophize this. I was just wondering: okay, then we have our language and our metalanguage, but how the son of a monkey do we formalize this? Then we need another metalanguage to describe this and that sucks. That's where I stopped 8-).
But there are a couple of things, you have your finite alphabet which I am fine with. But then you have your metalanguage to describe what you mean with "there exists" right? You have deduction rules, but you need some way to translate the result into something that does not look so scary. So then you use it to talk what the results mean you have. If it works this way I think I can understand how it goes.
FINAL EDIT : Prove that if $p^z|n^2-1$
$$p^{x-z}(p^{z}-1)=\dfrac{ n^2-1}{p^z}-3$$ doesn't hold for any chosen values of $p,x,n$ and $z$.
Here $p>3$ is an odd prime , $x=2y+z, \ \{\{x,y,z\}>0\} \in \mathbb{Z}$ . There $n$ is an even number.
If the above statement is prove it will lea...
